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If $z_1, z_2$ and $z_3, z_4$ are two pairs of conjugate complex numbers, then find $\arg \left(\frac{z_1}{z_4}\right)+\arg \left(\frac{z_2}{z_3}\right)$.
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Verified Answer
Let $z_1=r_1\left(\cos \theta_1+i \sin \theta_1\right)$,
Therefore, $z_2=\bar{z}_1=r_1\left(\cos \theta_1-i \sin \theta_1\right)$ $=r_1\left[\cos \left(-\theta_1\right)+\sin \left(-\theta_1\right)\right]$
Also, let $z_3=r_2\left(\cos \theta_2+i \sin \theta_2\right)$,
Therefore, $z_4=\bar{z}_3=r_2\left(\cos \theta_2-i \sin \theta_2\right)$
Then, $\arg \left(\frac{z_1}{z_4}\right)+\arg \left(\frac{z_2}{z_3}\right)$
$$
\begin{aligned}
&=\arg \left(z_1\right)-\arg \left(z_4\right)+\arg \left(z_2\right)-\arg \left(z_3\right) \\
&=\theta_1-\left(-\theta_2\right)+\left(-\theta_1\right)-\theta_2 \quad[\because \arg (z)=\theta] \\
&=\theta_1+\theta_2-\theta_1-\theta_2=0
\end{aligned}
$$
Therefore, $z_2=\bar{z}_1=r_1\left(\cos \theta_1-i \sin \theta_1\right)$ $=r_1\left[\cos \left(-\theta_1\right)+\sin \left(-\theta_1\right)\right]$
Also, let $z_3=r_2\left(\cos \theta_2+i \sin \theta_2\right)$,
Therefore, $z_4=\bar{z}_3=r_2\left(\cos \theta_2-i \sin \theta_2\right)$
Then, $\arg \left(\frac{z_1}{z_4}\right)+\arg \left(\frac{z_2}{z_3}\right)$
$$
\begin{aligned}
&=\arg \left(z_1\right)-\arg \left(z_4\right)+\arg \left(z_2\right)-\arg \left(z_3\right) \\
&=\theta_1-\left(-\theta_2\right)+\left(-\theta_1\right)-\theta_2 \quad[\because \arg (z)=\theta] \\
&=\theta_1+\theta_2-\theta_1-\theta_2=0
\end{aligned}
$$
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